BE Civil Engineering (IOE, TU) Basic Electrical Engineering (IOE, EE 401) Question Paper 2077 Nepal

(a) Nodal analysis

Take the bottom rail as reference ($0\,\text{V}$). Node A is held at $+42\,\text{V}$ and node C at $+10\,\text{V}$ by the ideal sources. Let the node-B voltage be $V_B$.

KCL at node B (sum of currents leaving B = 0):

$$\frac{V_B - 42}{6} + \frac{V_B - 10}{4} + \frac{V_B - 0}{12} = 0$$

Multiply through by the LCM, $12$:

$$2(V_B - 42) + 3(V_B - 10) + 1\,(V_B) = 0$$ $$2V_B - 84 + 3V_B - 30 + V_B = 0$$ $$6V_B - 114 = 0 \;\Rightarrow\; V_B = 19\,\text{V}$$

$V_B = 19\,\text{V}$

(b) Current through and power in $R_2$

$$I_{R_2} = \frac{V_B}{R_2} = \frac{19}{12} = 1.583\,\text{A (downward, B to reference)}$$ $$P_{R_2} = \frac{V_B^2}{R_2} = \frac{19^2}{12} = \frac{361}{12} = 30.08\,\text{W}$$

$I_{R_2} = 1.58\,\text{A},\quad P_{R_2} = 30.1\,\text{W}$

(c) Superposition check

Source 1 only (42 V active, 10 V replaced by short): Node C tied to reference, so $R_3$ ($4\,\Omega$) is now in parallel with $R_2$ ($12\,\Omega$):

$$R_2 \| R_3 = \frac{12 \times 4}{12 + 4} = 3\,\Omega$$

Voltage at B from the 42 V divider through $R_1 = 6\,\Omega$:

$$V_B' = 42 \times \frac{3}{6 + 3} = 42 \times \frac{3}{9} = 14\,\text{V}$$

Source 2 only (10 V active, 42 V replaced by short): Now $R_1$ ($6\,\Omega$) is in parallel with $R_2$ ($12\,\Omega$):

$$R_1 \| R_2 = \frac{6 \times 12}{6 + 12} = 4\,\Omega$$

Voltage at B from the 10 V divider through $R_3 = 4\,\Omega$:

$$V_B'' = 10 \times \frac{4}{4 + 4} = 10 \times \frac{1}{2} = 5\,\text{V}$$

Superpose: $V_B = V_B' + V_B'' = 14 + 5 = 19\,\text{V}$ — agrees with part (a). ✓

(a) Thevenin equivalent at A-B

Open-circuit voltage $V_{TH}$: with $R_L$ removed, $R_2$ carries the full source current. The current in the $R_1$–$R_2$ loop is

$$I = \frac{24}{R_1 + R_2} = \frac{24}{8 + 24} = \frac{24}{32} = 0.75\,\text{A}$$ $$V_{TH} = V_{R_2} = I \cdot R_2 = 0.75 \times 24 = 18\,\text{V}$$

Thevenin resistance $R_{TH}$: deactivate the source (short the 24 V). Then $R_1$ and $R_2$ appear in parallel across A-B:

$$R_{TH} = R_1 \| R_2 = \frac{8 \times 24}{8 + 24} = \frac{192}{32} = 6\,\Omega$$

$V_{TH} = 18\,\text{V},\quad R_{TH} = 6\,\Omega$

(b) Maximum power transfer theorem

Statement: A source network delivers maximum power to a load when the load resistance equals the Thevenin resistance of the source network as seen from the load terminals, i.e. $R_L = R_{TH}$.

$$\boxed{R_L = R_{TH} = 6\,\Omega}$$

(c) Maximum power and efficiency

With $R_L = R_{TH} = 6\,\Omega$, load current:

$$I_L = \frac{V_{TH}}{R_{TH} + R_L} = \frac{18}{6 + 6} = 1.5\,\text{A}$$ $$P_{L,\max} = I_L^2 R_L = (1.5)^2 \times 6 = 2.25 \times 6 = 13.5\,\text{W}$$

Alternatively $P_{\max} = V_{TH}^2 / (4 R_{TH}) = 324 / 24 = 13.5\,\text{W}$. ✓

Efficiency (load power ÷ total power delivered by Thevenin source):

$$\eta = \frac{P_L}{P_{total}} = \frac{R_L}{R_{TH} + R_L} = \frac{6}{12} = 0.5 = 50\%$$

$P_{L,\max} = 13.5\,\text{W},\quad \eta = 50\%$

Note: at maximum power transfer, efficiency is only 50% — the other half is lost in $R_{TH}$.

(a) Reactances, impedance, current, phase

$$X_L = 2\pi f L = 2\pi (50)(0.12) = 37.70\,\Omega$$ $$X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi (50)(80\times10^{-6})} = \frac{1}{0.025133} = 39.79\,\Omega$$

Net reactance $X = X_L - X_C = 37.70 - 39.79 = -2.09\,\Omega$ (capacitive).

$$Z = \sqrt{R^2 + X^2} = \sqrt{15^2 + (-2.09)^2} = \sqrt{225 + 4.37} = \sqrt{229.37} = 15.14\,\Omega$$ $$I = \frac{V}{Z} = \frac{230}{15.14} = 15.19\,\text{A}$$ $$\phi = \tan^{-1}\!\left(\frac{X}{R}\right) = \tan^{-1}\!\left(\frac{-2.09}{15}\right) = -7.94^\circ$$

Since $X_C > X_L$, the circuit is net capacitive and the current LEADS the voltage by about $7.9^\circ$.

$X_L = 37.7\,\Omega,\; X_C = 39.8\,\Omega,\; Z = 15.1\,\Omega,\; I = 15.2\,\text{A},\; \phi = 7.9^\circ$ leading

(b) Power and power factor

$$\text{pf} = \cos\phi = \cos(7.94^\circ) = 0.990\ \text{(leading)}$$ $$P = VI\cos\phi = 230 \times 15.19 \times 0.990 = 3459\,\text{W} \approx 3.46\,\text{kW}$$ $$Q = VI\sin\phi = 230 \times 15.19 \times \sin(7.94^\circ) = 230 \times 15.19 \times 0.1382 = 482.8\,\text{VAR (capacitive)}$$ $$S = VI = 230 \times 15.19 = 3494\,\text{VA} \approx 3.49\,\text{kVA}$$

Check: $\sqrt{P^2+Q^2} = \sqrt{3459^2 + 483^2} = \sqrt{11.96\times10^6 + 0.233\times10^6} = 3493\,\text{VA}$ ✓

$P = 3.46\,\text{kW},\; Q = 0.48\,\text{kVAR (leading)},\; S = 3.49\,\text{kVA},\; \text{pf} = 0.99$ leading

(c) Resonance

$$f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.12 \times 80\times10^{-6}}} = \frac{1}{2\pi\sqrt{9.6\times10^{-6}}}$$ $$\sqrt{9.6\times10^{-6}} = 3.098\times10^{-3}\,\text{s}$$ $$f_0 = \frac{1}{2\pi (3.098\times10^{-3})} = \frac{1}{0.01947} = 51.37\,\text{Hz}$$

At resonance $X_L = X_C$, so $Z = R = 15\,\Omega$ and the current is maximum:

$$I_0 = \frac{V}{R} = \frac{230}{15} = 15.33\,\text{A}$$

$f_0 = 51.4\,\text{Hz},\quad I_0 = 15.3\,\text{A}$

(a) Phase voltage, currents, power factor (star)

For a star connection $V_{ph} = V_L/\sqrt{3}$:

$$V_{ph} = \frac{400}{\sqrt{3}} = 230.9\,\text{V}$$

Per-phase impedance:

$$Z_{ph} = \sqrt{R^2 + X_L^2} = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25\,\Omega$$ $$I_{ph} = \frac{V_{ph}}{Z_{ph}} = \frac{230.9}{25} = 9.24\,\text{A}$$

In star, line current = phase current: $I_L = I_{ph} = 9.24\,\text{A}$.

Power factor:

$$\cos\phi = \frac{R}{Z} = \frac{20}{25} = 0.8\ \text{(lagging)}$$

$V_{ph} = 230.9\,\text{V},\; I_{ph} = I_L = 9.24\,\text{A},\; \text{pf} = 0.8$ lag

(b) Powers (star)

$$P = \sqrt{3}\,V_L I_L \cos\phi = \sqrt{3}\,(400)(9.24)(0.8) = 1.7321 \times 400 \times 9.24 \times 0.8 = 5121\,\text{W}$$ $$Q = \sqrt{3}\,V_L I_L \sin\phi = \sqrt{3}\,(400)(9.24)(0.6) = 3841\,\text{VAR}$$ $$S = \sqrt{3}\,V_L I_L = 1.7321 \times 400 \times 9.24 = 6401\,\text{VA}$$

Check per-phase: $P = 3 I_{ph}^2 R = 3 (9.24)^2 (20) = 3 \times 85.38 \times 20 = 5123\,\text{W}$ ✓ (rounding).

$P \approx 5.12\,\text{kW},\; Q \approx 3.84\,\text{kVAR},\; S \approx 6.40\,\text{kVA}$

(c) Same impedance reconnected in delta

In delta the full line voltage appears across each phase: $V_{ph} = V_L = 400\,\text{V}$.

$$I_{ph,\Delta} = \frac{V_{ph}}{Z_{ph}} = \frac{400}{25} = 16\,\text{A}$$ $$I_{L,\Delta} = \sqrt{3}\,I_{ph,\Delta} = 1.7321 \times 16 = 27.71\,\text{A}$$

$I_{L,\Delta} = 27.7\,\text{A}$ — three times the star line current ($9.24 \times 3 = 27.7\,\text{A}$), illustrating the $P_\Delta = 3 P_Y$ relationship.

(a) Full-load efficiency at 0.8 pf lagging

Full-load output power:

$$P_{out} = S \times \text{pf} = 25{,}000 \times 0.8 = 20{,}000\,\text{W}$$

Total losses at full load = iron loss + full-load copper loss:

$$P_{loss} = 250 + 400 = 650\,\text{W}$$ $$\eta = \frac{P_{out}}{P_{out} + P_{loss}} = \frac{20{,}000}{20{,}000 + 650} = \frac{20{,}000}{20{,}650} = 0.9685 = 96.85\%$$

$\eta_{FL} = 96.85\%$

(b) Load for maximum efficiency

Maximum efficiency occurs when copper loss = iron loss. If $x$ is the fraction of full load, copper loss $= x^2 P_{cu,FL}$:

$$x^2 P_{cu,FL} = P_{iron} \;\Rightarrow\; x = \sqrt{\frac{P_{iron}}{P_{cu,FL}}} = \sqrt{\frac{250}{400}} = \sqrt{0.625} = 0.7906$$

So maximum efficiency is at $x = 0.79$, i.e. $79.1\%$ of full load.

Output at this load = $0.7906 \times 25{,}000 \times 0.8 = 15{,}811\,\text{W}$. Losses = $2 \times P_{iron} = 2 \times 250 = 500\,\text{W}$ (copper = iron at this point).

$$\eta_{\max} = \frac{15{,}811}{15{,}811 + 500} = \frac{15{,}811}{16{,}311} = 0.9693 = 96.93\%$$

$x = 0.791$ of full load, $\eta_{\max} = 96.93\%$

(c) Voltage regulation at full load, 0.8 pf lagging

Work on the LV (200 V) side. Full-load secondary current:

$$I_2 = \frac{S}{V_2} = \frac{25{,}000}{200} = 125\,\text{A}$$

For a lagging pf, $\cos\phi = 0.8$, $\sin\phi = 0.6$. Per-unit-free regulation formula:

$$\%\text{Reg} = \frac{I_2(R_{02}\cos\phi + X_{02}\sin\phi)}{V_2}\times 100$$ $$= \frac{125\,(0.04\times0.8 + 0.06\times0.6)}{200}\times 100$$ $$= \frac{125\,(0.032 + 0.036)}{200}\times 100 = \frac{125 \times 0.068}{200}\times 100 = \frac{8.5}{200}\times 100 = 4.25\%$$

Voltage regulation $= 4.25\%$

(a) Equivalent capacitance

Series combination of $C_1$ and $C_2$:

$$C_{12} = \frac{C_1 C_2}{C_1 + C_2} = \frac{4 \times 6}{4 + 6} = \frac{24}{10} = 2.4\,\mu\text{F}$$

This in parallel with $C_3$:

$$C_{eq} = C_{12} + C_3 = 2.4 + 12 = 14.4\,\mu\text{F}$$

$C_{eq} = 14.4\,\mu\text{F}$

(b) Total charge and energy

$$Q_{total} = C_{eq} V = 14.4\times10^{-6} \times 100 = 1.44\times10^{-3}\,\text{C} = 1.44\,\text{mC}$$ $$E = \tfrac{1}{2} C_{eq} V^2 = \tfrac{1}{2}(14.4\times10^{-6})(100)^2 = \tfrac{1}{2}(14.4\times10^{-6})(10{,}000) = 0.072\,\text{J} = 72\,\text{mJ}$$

$Q_{total} = 1.44\,\text{mC},\quad E = 72\,\text{mJ}$

(c) Voltage across $C_1$

The series branch ($C_{12} = 2.4\,\mu\text{F}$) has the full $100\,\text{V}$ across it. Its charge:

$$Q_{12} = C_{12} V = 2.4\times10^{-6} \times 100 = 2.4\times10^{-4}\,\text{C}$$

In a series branch each capacitor carries the same charge $Q_{12}$, so:

$$V_{C_1} = \frac{Q_{12}}{C_1} = \frac{2.4\times10^{-4}}{4\times10^{-6}} = 60\,\text{V}$$

(Check: $V_{C_2} = Q_{12}/C_2 = 2.4\times10^{-4}/6\times10^{-6} = 40\,\text{V}$; $60 + 40 = 100\,\text{V}$ ✓)

$V_{C_1} = 60\,\text{V}$

(a) MMF and reluctance

Magnetomotive force:

$$\mathcal{F} = N I = 400 \times 2 = 800\,\text{A-turns (At)}$$

Reluctance of the iron core:

$$\mathcal{S} = \frac{l}{\mu_0 \mu_r A} = \frac{0.5}{(4\pi\times10^{-7})(800)(5\times10^{-4})}$$

Denominator: $(4\pi\times10^{-7}) = 1.2566\times10^{-6}$; $\times 800 = 1.00531\times10^{-3}$; $\times 5\times10^{-4} = 5.0265\times10^{-7}$.

$$\mathcal{S} = \frac{0.5}{5.0265\times10^{-7}} = 9.947\times10^{5}\,\text{At/Wb}$$

$\mathcal{F} = 800\,\text{At},\quad \mathcal{S} = 9.95\times10^{5}\,\text{At/Wb}$

(b) Flux and flux density

$$\Phi = \frac{\mathcal{F}}{\mathcal{S}} = \frac{800}{9.947\times10^{5}} = 8.04\times10^{-4}\,\text{Wb} = 0.804\,\text{mWb}$$ $$B = \frac{\Phi}{A} = \frac{8.04\times10^{-4}}{5\times10^{-4}} = 1.608\,\text{T}$$

Check via $B = \mu_0\mu_r H$, $H = NI/l = 800/0.5 = 1600\,\text{At/m}$: $B = 1.2566\times10^{-6}\times800\times1600 = 1.608\,\text{T}$ ✓

$\Phi = 0.804\,\text{mWb},\quad B = 1.61\,\text{T}$

Setup

Let $I_1$ = current from Battery 1, $I_2$ = current from Battery 2. Load current $I_L = I_1 + I_2$ (KCL at the load node). Load voltage $V_L = I_L R_L$.

KVL for each battery loop (EMF = internal drop + load voltage):

$$12 = 0.3\,I_1 + V_L \quad\text{(1)}$$ $$10 = 0.2\,I_2 + V_L \quad\text{(2)}$$

with $V_L = 2(I_1 + I_2)$.

Solve

From (1): $I_1 = \dfrac{12 - V_L}{0.3}$. From (2): $I_2 = \dfrac{10 - V_L}{0.2}$.

Substitute into $V_L = 2(I_1 + I_2)$:

$$V_L = 2\left(\frac{12 - V_L}{0.3} + \frac{10 - V_L}{0.2}\right)$$ $$V_L = 2\big(40 - 3.333V_L + 50 - 5V_L\big) = 2(90 - 8.333V_L)$$ $$V_L = 180 - 16.667V_L$$ $$17.667\,V_L = 180 \;\Rightarrow\; V_L = 10.19\,\text{V}$$

Then:

$$I_1 = \frac{12 - 10.19}{0.3} = \frac{1.81}{0.3} = 6.04\,\text{A}$$ $$I_2 = \frac{10 - 10.19}{0.2} = \frac{-0.19}{0.2} = -0.94\,\text{A}$$

The negative sign means Battery 2 is actually being charged (current flows into it).

Check: $I_L = 6.04 + (-0.94) = 5.09\,\text{A}$; $V_L = I_L R_L = 5.09 \times 2 = 10.19\,\text{V}$ ✓

$I_1 = 6.04\,\text{A}$ (discharging), $I_2 = -0.94\,\text{A}$ (i.e. 0.94 A charging), $V_L = 10.19\,\text{V}$

(a) Basic quantities

From $i(t) = I_m \sin(\omega t)$: peak $I_m = 14.14\,\text{A}$, $\omega = 314\,\text{rad/s}$.

$$I_{rms} = \frac{I_m}{\sqrt{2}} = \frac{14.14}{1.4142} = 10.0\,\text{A}$$ $$I_{avg} = \frac{2 I_m}{\pi} = \frac{2 \times 14.14}{3.1416} = 9.003\,\text{A} \approx 9.0\,\text{A}$$ $$f = \frac{\omega}{2\pi} = \frac{314}{6.2832} = 49.97\,\text{Hz} \approx 50\,\text{Hz}$$ $$T = \frac{1}{f} = \frac{1}{50} = 0.02\,\text{s} = 20\,\text{ms}$$

$I_m = 14.14\,\text{A},\; I_{rms} = 10\,\text{A},\; I_{avg} = 9.0\,\text{A},\; f = 50\,\text{Hz},\; T = 20\,\text{ms}$

(b) Form factor and peak factor

Form factor = RMS value / average value:

$$k_f = \frac{I_{rms}}{I_{avg}} = \frac{10}{9.0} = 1.11$$

Peak (crest) factor = peak value / RMS value:

$$k_p = \frac{I_m}{I_{rms}} = \frac{14.14}{10} = 1.414$$

(These are the standard values $1.11$ and $1.414$ for a pure sinusoid.)

$k_f = 1.11,\quad k_p = 1.414$

(c) Instantaneous value at $t = 2.5\,\text{ms}$

$$\omega t = 314 \times 2.5\times10^{-3} = 0.785\,\text{rad}\;(=45^\circ)$$ $$i = 14.14 \sin(0.785) = 14.14 \times 0.7071 = 10.0\,\text{A}$$

$i(2.5\,\text{ms}) = 10.0\,\text{A}$

(a) Field and armature currents

Field (shunt) current:

$$I_{sh} = \frac{V}{R_{sh}} = \frac{220}{110} = 2\,\text{A}$$

Armature current (line current minus field current):

$$I_a = I_L - I_{sh} = 22 - 2 = 20\,\text{A}$$

$I_{sh} = 2\,\text{A},\quad I_a = 20\,\text{A}$

(b) Back EMF

For a motor, $E_b = V - I_a R_a$:

$$E_b = 220 - (20)(0.4) = 220 - 8 = 212\,\text{V}$$

$E_b = 212\,\text{V}$

(c) Power developed by the armature

Gross mechanical (developed) power = back EMF × armature current:

$$P_{dev} = E_b I_a = 212 \times 20 = 4240\,\text{W} = 4.24\,\text{kW}$$

(For reference, total input $= V I_L = 220 \times 22 = 4840\,\text{W}$; armature copper loss $= I_a^2 R_a = 400 \times 0.4 = 160\,\text{W}$; field loss $= V I_{sh} = 440\,\text{W}$; $4840 - 160 - 440 = 4240\,\text{W}$ ✓)

$P_{dev} = 4.24\,\text{kW}$

(a) Load powers

Apparent power:

$$S = V I = 230 \times 30 = 6900\,\text{VA} = 6.9\,\text{kVA}$$

Active power (pf = 0.7):

$$P = S\cos\phi = 6900 \times 0.7 = 4830\,\text{W} = 4.83\,\text{kW}$$

Reactive power ($\sin\phi = \sqrt{1 - 0.7^2} = \sqrt{0.51} = 0.7141$):

$$Q_1 = S\sin\phi = 6900 \times 0.7141 = 4927\,\text{VAR (lagging)}$$

$P = 4.83\,\text{kW},\; Q_1 = 4.93\,\text{kVAR},\; S = 6.9\,\text{kVA}$

(b) Capacitor for pf correction to 0.95 lagging

Active power stays $P = 4830\,\text{W}$. New phase angle:

$$\phi_2 = \cos^{-1}(0.95) = 18.19^\circ,\quad \tan\phi_2 = 0.3287$$

Old angle: $\phi_1 = \cos^{-1}(0.7) = 45.57^\circ,\quad \tan\phi_1 = 1.0202$.

Required reactive power supplied by the capacitor:

$$Q_C = P(\tan\phi_1 - \tan\phi_2) = 4830\,(1.0202 - 0.3287) = 4830 \times 0.6915 = 3340\,\text{VAR}$$

Capacitor reactive power $Q_C = V^2 / X_C = V^2 \cdot 2\pi f C$, so:

$$C = \frac{Q_C}{2\pi f V^2} = \frac{3340}{2\pi (50)(230)^2} = \frac{3340}{2\pi (50)(52{,}900)}$$ $$= \frac{3340}{16{,}618{,}540} = 2.010\times10^{-4}\,\text{F} = 201\,\mu\text{F}$$

$Q_C = 3.34\,\text{kVAR},\quad C \approx 201\,\mu\text{F}$